poj3415之長度不小於k的公共子串個數
阿新 • • 發佈:2018-12-24
Common Substrings
T(i, k)=TiTi+1...Ti+k-1, 1≤i≤i+k-1≤|T|.
S = {(i, j, k) | k≥K, A(i, k)=B(j, k)}.
| for specific A, B and K.
Time Limit: 5000MS | Memory Limit: 65536K |
Total Submissions: 6254 | Accepted: 2069 |
Description
A substring of a string T is defined as:
Given two strings A, B and one integer K, we define S, a set of triples (i, j, k):
You are to give the value of |S
Input
The input file contains several blocks of data. For each block, the first line contains one integer K, followed by two lines containing strings A and B, respectively. The input file is ended by K=0.
1 ≤ |A|, |B| ≤ 105
1 ≤ K ≤ min{|A|, |B|}
Characters of A and B are all Latin letters.
Output
For each case, output an integer |S|.
Sample Input
2 aababaa abaabaa 1 xx xx 0
Sample Output
22 5
#include<iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <queue> #include <algorithm> #include <map> #include <iomanip> #define INF 99999999 typedef long long LL; using namespace std; const int MAX=2*100000+10; int *rank,r[MAX],sa[MAX],height[MAX]; int wa[MAX],wb[MAX],wm[MAX]; char s[MAX]; bool cmp(int *r,int a,int b,int l){ return r[a] == r[b] && r[a+l] == r[b+l]; } void makesa(int *r,int *sa,int n,int m){ int *x=wa,*y=wb,*t; for(int i=0;i<m;++i)wm[i]=0; for(int i=0;i<n;++i)wm[x[i]=r[i]]++; for(int i=1;i<m;++i)wm[i]+=wm[i-1]; for(int i=n-1;i>=0;--i)sa[--wm[x[i]]]=i; for(int i=0,j=1,p=0;p<n;j=j*2,m=p){ for(p=0,i=n-j;i<n;++i)y[p++]=i; for(i=0;i<n;++i)if(sa[i]>=j)y[p++]=sa[i]-j; for(i=0;i<m;++i)wm[i]=0; for(i=0;i<n;++i)wm[x[y[i]]]++; for(i=1;i<m;++i)wm[i]+=wm[i-1]; for(i=n-1;i>=0;--i)sa[--wm[x[y[i]]]]=y[i]; for(t=x,x=y,y=t,i=p=1,x[sa[0]]=0;i<n;++i){ x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++; } } rank=x; } void calheight(int *r,int *sa,int n){ for(int i=0,j=0,k=0;i<n;height[rank[i++]]=k){ for(k?--k:0,j=sa[rank[i]-1];r[i+k] == r[j+k];++k); } } LL calculate(int n,int len,int k){ int *mark=wb,*ans=wm,Top=0;//num[1],num[2]分別表示字串A,B,suffix(0~i-1)和suffix(i)的最長公共子串>=k的總個數 LL sum=0,num[3]={0}; for(int i=1;i<=n;++i){ if(height[i]<k){ Top=num[1]=num[2]=0; }else{ for(int size=Top;size && ans[size]>height[i]-k+1;--size){//維護單調棧,ans記錄的是suffix(j)和suffix(i-1)>=k的最長公共子串的個數,個數越多表示height[j]越大 num[mark[size]]+=height[i]-k+1-ans[size];//suffix(j)和suffix(i)>=k的最長公共子串只能是長度為k~height[i],所以需要減去(ans[size]-(height[i]-k+1)) ans[size]=height[i]-k+1;//更新個數(更新單調棧,使棧裡面元素非遞減) } ans[++Top]=height[i]-k+1; if(sa[i-1]<len)mark[Top]=1;//由於num新增加的結果是suffix(i-1)和suffix(i)的結果,所以是判斷sa[i-1] if(sa[i-1]>len)mark[Top]=2; num[mark[Top]]+=height[i]-k+1;//增加由suffix(i-1)和suffix(i)產生的結果 if(sa[i]<len)sum+=num[2];//表示和suffix(i)產生的結果新增加B串的suffix(0~i-1)和suffix(i)>=k的個數 if(sa[i]>len)sum+=num[1];//表示和suffix(i)產生的結果新增加A串的suffix(0~i-1)和suffix(i)>=k的個數 } } return sum; } int main(){ int k,n,len; while(~scanf("%d",&k),k){ scanf("%s",s); for(n=0;s[n] != '\0';++n)r[n]=s[n]; r[len=n]='#'; scanf("%s",s+n+1); for(++n;s[n] != '\0';++n)r[n]=s[n]; r[n]=0; makesa(r,sa,n+1,256); calheight(r,sa,n); cout<<calculate(n,len,k)<<endl; } return 0; }